If a 10.0 kg object is on a surface that is inclined 30o and the coefficient of static friction is 0.65, what is the force of st
1 answer:
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Answer:
Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.
Explanation:
The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is
K= constant depends on the system of units used.
W= weight =485 g
D= density =7.9 g/cm³
A = exposed specimen area =100 in² =6.452 cm²
K=534 to give CPR in mpy
K=87.6 to give CPR in mm/yr
mpy
=37.4mpy
mm/yr
=0.952 mm/yr
Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.
Answer:
μ = 0.692
Explanation:
In order to solve this problem, we must make a free body diagram and include the respective forces acting on the body. Similarly, deduce the respective equations according to the conditions of the problem and the directions of the forces.
Attached is an image with the respective forces:
A summation of forces on the Y-axis is performed equal to zero, in order to determine the normal force N. this summation is equal to zero since there is no movement on the Y-axis.
Since the body moves at a constant speed, there is no acceleration so the sum of forces on the X-axis must be equal to zero.
The frictional force is defined as the product of the coefficient of friction by the normal force. In this way, we can calculate the coefficient of friction.
The process of solving this problem can be seen in the attached image.
